Solving equations involving exponents and logarithms

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Transcript of Solving equations involving exponents and logarithms

  • Solving equations involving exponents and logarithms

  • Lets review some terms.

    When we write log5 125

    5 is called the base125 is called the argument

  • Logarithmic form of 52 = 25 is

    log525 = 2

  • For all the laws a, M and N > 0a 1r is any real

  • Remember ln and logln is a short cut for logelog means log10

  • Log laws

  • If your variable is in an exponent or in the argument of a logarithmFind the pattern your equation resembles

  • If your variable is in an exponent or in the argument of a logarithmFind the pattern Fit your equation to match the patternSwitch to the equivalent formSolve the resultCheck (be sure you remember the domain of a log)

  • log(2x) = 3

  • It fits log(2x) = 3

  • Switch log(2x) = 3103=2xDid you remember that log(2x) means log10(2x)?

  • Divide by 2 log(2x) = 3103=2x500 = x

  • ln(x+3) = ln(-7x)

  • ln(x+3) = ln(-7x) It fits

  • ln(x+3) = ln(-7x)Switch

  • ln(x+3) = ln(-7x)x + 3 = -7x

    Switch

  • ln(x+3) = ln(-7x)x + 3 = -7xx = -

    Solve the result(and check)

  • ln(x) + ln(3) = ln(12)

  • ln(x) + ln(3) = ln(12)x + 3 = 12

  • ln(x) + ln(3) = ln(12)x + 3 = 12 Oh NO!!! Thats wrong!

  • You need to use log lawsln(x) + ln(3) = ln(12) ln(3x) = ln (12)

  • Switchln(x) + ln(3) = ln(12)ln(3x) = ln (12)3x = 12

  • ln(x) + ln(3) = ln(12)ln(3x) = ln (12)3x = 12x = 4Solve the result

  • log3(x+2) + 4 = 9

  • It will fit log3(x+2) + 4 = 9

  • Subtract 4 to make it fit log3(x+2) + 4 = 9log3(x+2) = 5

  • Switch log3(x+2) + 4 = 9log3(x+2) = 5

  • Switch log3(x+2) + 4 = 9log3(x+2) = 535 = x + 2

  • Solve the resultlog3(x+2) + 4 = 9log3(x+2) = 535 = x + 2 x = 241

  • 5(10x) = 19.45

  • Divide by 5 to fit5(10x) = 19.4510x = 3.91

  • Switch 5(10x) = 19.4510x = 3.91

  • Switch 5(10x) = 19.4510x = 3.91log(3.91) = x

  • Exact log(3.91) Approx 0.5925(10x) = 19.4510x = 3.91log(3.91) = x 0.592

  • 2 log3(x) = 8

  • It will fit 2 log3(x) = 8

  • Divide by 2 to fit 2 log3(x) = 8log3(x) = 4

  • Switch 2 log3(x) = 8log3(x) = 4

  • Switch 2 log3(x) = 8log3(x) = 434=x

  • Then Simplify2 log3(x) = 8log3(x) = 434=x x = 81

  • log2(x-1) + log2(x-1) = 3

  • Need to use a log lawlog2(x-1) + log2(x-1) = 3

  • log2(x-1) + log2(x+1) = 3log2{(x-1)(x+1)} = 3

  • Switchlog2(x-1) + log2(x+1) = 3log2{(x-1)(x+1)} = 3

  • Switchlog2(x-1) + log2(x+1) = 3log2{(x-1)(x+1)} = 323=(x-1)(x+1)

  • and finishlog2(x-1) + log2(x+1) = 3log2{(x-1)(x+1)} = 323=(x-1)(x+1) = x2 -1x = +3 or -3

  • But -3 does not check!log2(x-1) + log2(x+1) = 3log2{(x-1)(x+1)} = 323=(x-1)(x+1) = x2 -1x = +3 or -3

  • Exclude -3 (it would cause you to have a negative argument)log2(x-1) + log2(x+1) = 3log2{(x-1)(x+1)} = 323=(x-1)(x+1) = x2 -1x = +3 or -3

  • Theres more than one way to do this

  • Can you find why each step is valid?

  • rules of exponents multiply both sides by 2

    - 3 to get exact answerApproximate answer

  • Heres another way to solve the same equation.

  • exclude 2nd resultSquare both sidesSimplify

  • 52x - 5x 12 = 0

  • Factor it. Think of y2 - y-12=052x - 5x 12 = 0(5x 4)(5x + 3) = 0

  • Set each factor = 052x - 5x 12 = 0(5x 4)(5x + 3) = 05x 4 = 0 or 5x + 3 = 0

  • Solve first factors equationSolve 5x 4 = 0 5x = 4log54 = x

  • Solve other factors equationSolve 5x + 3 = 0 5x = -3log5(-3) = x

  • Oops, we cannot have a negative argumentSolve 5x + 3 = 0 5x = -3log5(-3) = x

  • Exclude this solution.Only the other factors solution worksSolve 5x + 3 = 0 5x = -3log5(-3) = x

  • 4x+2 = 5x

  • If M = N then ln M = ln N

    4x+2 = 5x

  • If M = N then ln M = ln N4x+2 = 5x ln(4x+2) = ln(5x )

  • 4x+2 = 5x ln(4x+2) = ln(5x)

  • 4x+2 = 5xln(4x+2) = ln(5x)(x+2)ln(4) =(x) ln(5 )

  • Distribute 4x+2 = 5xln(4x+2) = ln(5x)(x+2)ln(4) =(x) ln(5 )x ln (4) + 2 ln(4) = x ln (5)

  • Get x terms on one side 4x+2 = 5x ln(4x+2) = ln(5x)(x+2)ln(4) =(x) ln(5 )x ln (4) + 2 ln(4) = x ln (5)x ln (4) - x ln (5)= 2 ln(4)

  • Factor out x 4x+2 = 5xln(4x+2) = ln(5x)(x+2)ln(4) =(x) ln(5 )x ln (4) + 2 ln(4) = x ln (5)x ln (4) - x ln (5)= 2 ln(4) x [ln (4) - ln (5)]= 2 ln(4)

  • Divide by numerical coefficient 4x+2 = 5xln(4x+2) = ln(5x)(x+2)ln(4) =(x) ln(5 )x ln (4) + 2 ln(4) = x ln (5)x ln (4) - x ln (5)= 2 ln(4) x [ln (4) - ln (5)]= 2 ln(4)